def show_topk(self, k):
    rank = self.result.argsort() # 排序 能看那个最大概率的索引是否在最后最大位置上
    hit_top_k = [l in rank[i, -k:] for i, l in enumerate(self.label)]
    accuracy = sum(hit_top_k) * 1.0 / len(hit_top_k)
    self.io.print_log('\tTop{}: {:.2f}%'.format(k, 100 * accuracy))